YES

Problem:
 *(i(x),x) -> 1()
 *(1(),y) -> y
 *(x,0()) -> 0()
 *(*(x,y),z) -> *(x,*(y,z))

Proof:
 DP Processor:
  DPs:
   *#(*(x,y),z) -> *#(y,z)
   *#(*(x,y),z) -> *#(x,*(y,z))
  TRS:
   *(i(x),x) -> 1()
   *(1(),y) -> y
   *(x,0()) -> 0()
   *(*(x,y),z) -> *(x,*(y,z))
  KBO Processor:
   argument filtering:
    pi(i) = 0
    pi(*) = [0,1]
    pi(1) = []
    pi(0) = []
    pi(*#) = [0,1]
   usable rules:
    *(i(x),x) -> 1()
    *(1(),y) -> y
    *(x,0()) -> 0()
    *(*(x,y),z) -> *(x,*(y,z))
   weight function:
    w0 = 1
    w(*#) = w(0) = w(1) = w(i) = 1
    w(*) = 0
   precedence:
    *# ~ 0 ~ 1 ~ * ~ i
   problem:
    DPs:
     
    TRS:
     *(i(x),x) -> 1()
     *(1(),y) -> y
     *(x,0()) -> 0()
     *(*(x,y),z) -> *(x,*(y,z))
   Qed